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| david.cawkwell@tesco.net 2006-06-20, 5:25 pm |
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I have a water heater.
The heater has 3 elements and is rated at 4.5kw.
Therefore each element is 1.5kw.
If I measure the resistance of the heating element I get about 97Ohms.
Therefore on 240 volts AC rms I should have 240/97=2.5 Amps
Now P=I*I*R so power = 2.5*2.5*97=606watts. Total is 3*606 so 1.8kw.
Are my calculations wrong or why am I not getting 4.5kw?
| |
| Palindr☻me 2006-06-20, 5:25 pm |
| david.cawkwell@tesco.net wrote:
>
> I have a water heater.
>
> The heater has 3 elements and is rated at 4.5kw.
>
> Therefore each element is 1.5kw.
>
> If I measure the resistance of the heating element I get about 97Ohms.
>
> Therefore on 240 volts AC rms I should have 240/97=2.5 Amps
>
> Now P=I*I*R so power = 2.5*2.5*97=606watts. Total is 3*606 so 1.8kw.
>
> Are my calculations wrong or why am I not getting 4.5kw?
>
At a rough guess - your resistance reading is in error.
--
Sue
| |
| John C 2006-06-20, 8:25 pm |
| resistance is relative.
You need to include impedance of the total cct in parallel.
p=ir
4500=970*I
I=4500/970
4.6amps......
1/970 etc
total resistance is 323 ohms
more confusion?
"Palindr?me" <me9@privacy.net> wrote in message
news:129gt19bm2g4988@corp.supernews.com...
> david.cawkwell@tesco.net wrote:
>
> At a rough guess - your resistance reading is in error.
>
> --
> Sue
| |
| sQuick 2006-06-20, 8:25 pm |
|
<david.cawkwell@tesco.net> wrote in message
news:1150841097.646017.251120@r2g2000cwb.googlegroups.com...
>
>
> I have a water heater.
>
> The heater has 3 elements and is rated at 4.5kw.
>
> Therefore each element is 1.5kw.
>
> If I measure the resistance of the heating element I get about 97Ohms.
>
> Therefore on 240 volts AC rms I should have 240/97=2.5 Amps
>
> Now P=I*I*R so power = 2.5*2.5*97=606watts. Total is 3*606 so 1.8kw.
>
> Are my calculations wrong or why am I not getting 4.5kw?
>
Single phase or three phase star/y connected ?
| |
| Ryan Evans 2006-06-20, 8:25 pm |
| Most home water heaters have 2 elements - and they do not operate
simultaneously.
Each one is 4.5 kW by itself. -- 240^2/4500 = 12.8 ohms
RE
<david.cawkwell@tesco.net> wrote in message
news:1150841097.646017.251120@r2g2000cwb.googlegroups.com...
>
>
> I have a water heater.
>
> The heater has 3 elements and is rated at 4.5kw.
>
> Therefore each element is 1.5kw.
>
> If I measure the resistance of the heating element I get about 97Ohms.
>
> Therefore on 240 volts AC rms I should have 240/97=2.5 Amps
>
> Now P=I*I*R so power = 2.5*2.5*97=606watts. Total is 3*606 so 1.8kw.
>
> Are my calculations wrong or why am I not getting 4.5kw?
>
| |
| Peter Pan 2006-06-21, 3:25 am |
| Don't forget that resistance drop their resistance as the temperature
rises. There are tables you can consult, depending on the wire material
and the temperature it reaches.
In article <kq-dnYaGrfVbHgXZnZ2dnUVZ_u2dnZ2d@comcast.com>, Ryan Evans
<ryanevans@comcast.net> wrote:
> Most home water heaters have 2 elements - and they do not operate
> simultaneously.
> Each one is 4.5 kW by itself. -- 240^2/4500 = 12.8 ohms
>
> RE
>
>
> <david.cawkwell@tesco.net> wrote in message
> news:1150841097.646017.251120@r2g2000cwb.googlegroups.com...
>
>
| |
|
| david.cawkwell@tesco.net wrote:
>
> I have a water heater.
>
> The heater has 3 elements and is rated at 4.5kw.
>
> Therefore each element is 1.5kw.
To get 1.5 kW out of an element at 240 volts
requires 6.25 amps. To draw 6.25 amps at 240
volts requires 38.4 ohms. You have a number
wrong, somewhere, likely the resistance measurement.
Ed
>
> If I measure the resistance of the heating element I get about 97Ohms.
>
> Therefore on 240 volts AC rms I should have 240/97=2.5 Amps
>
> Now P=I*I*R so power = 2.5*2.5*97=606watts. Total is 3*606 so 1.8kw.
>
> Are my calculations wrong or why am I not getting 4.5kw?
>
| |
| gfretwell@aol.com 2006-06-21, 3:25 am |
| On Wed, 21 Jun 2006 03:37:24 GMT, ehsjr <ehsjr@bellatlantic.net>
wrote:
>To get 1.5 kW out of an element at 240 volts
>requires 6.25 amps. To draw 6.25 amps at 240
>volts requires 38.4 ohms. You have a number
>wrong, somewhere, likely the resistance measurement.
>Ed
I bet he dropped a decimal point and these are 9.xx ohms. A water
heater usually just uses one elememt at a time so they are all rated
at the heater rating.
| |
| gfretwell@aol.com 2006-06-21, 3:25 am |
| On Tue, 20 Jun 2006 21:22:45 -0500, Peter Pan <pR_pMaVaEn4SPM@mac.com>
wrote:
>Don't forget that resistance drop their resistance as the temperature
>rises.
You have that backward or a lamp filiment would explode. Nichrome wire
increases resistance as it heats up. I imagine he is really reading
9.7 ohms cold and that goes to 12.8 hot. Do the math, that works out
to 4.5kw @ 240v. Water heaters usually only use one element at a time.
| |
| david.cawkwell@tesco.net 2006-06-21, 1:25 pm |
| No I measured all 3 on the 200ohms scale of the meter.
All came back with 97 ohms.
I'll go with the idea that the resistance will drop as the elements get
warm.
Can't be 9.7 ohms as the current would be much too high.
gfretwell@aol.com wrote:
> On Wed, 21 Jun 2006 03:37:24 GMT, ehsjr <ehsjr@bellatlantic.net>
> wrote:
>
> I bet he dropped a decimal point and these are 9.xx ohms. A water
> heater usually just uses one elememt at a time so they are all rated
> at the heater rating.
| |
| david.cawkwell@tesco.net 2006-06-21, 1:25 pm |
| This is a spa bath heater.
I think all the elements are on at once. I have traced the wiring back
the to contactor
so they all come on at once.
gfretwell@aol.com wrote:
> On Wed, 21 Jun 2006 03:37:24 GMT, ehsjr <ehsjr@bellatlantic.net>
> wrote:
>
> I bet he dropped a decimal point and these are 9.xx ohms. A water
> heater usually just uses one elememt at a time so they are all rated
> at the heater rating.
| |
|
|
<david.cawkwell@tesco.net> wrote in message
news:1150841097.646017.251120@r2g2000cwb.googlegroups.com...
>
>
> I have a water heater.
>
> The heater has 3 elements and is rated at 4.5kw.
>
> Therefore each element is 1.5kw.
>
> If I measure the resistance of the heating element I get about 97Ohms.
>
> Therefore on 240 volts AC rms I should have 240/97=2.5 Amps
>
> Now P=I*I*R so power = 2.5*2.5*97=606watts. Total is 3*606 so 1.8kw.
>
> Are my calculations wrong or why am I not getting 4.5kw?
>
probably your measurement - the heating element changes (increases)
resistance as it heats up.
| |
|
| david.cawkwell@tesco.net wrote:
> No I measured all 3 on the 200ohms scale of the meter.
> All came back with 97 ohms.
>
> I'll go with the idea that the resistance will drop as the elements get
> warm.
It may be time to replace the battery in your meter.
In fact, get two new batteries - one for the meter
and the other for the paragraph below. Proceed with
the paragraph, after replacing the battery in your
meter. A weak battery can cause erroneous resistance
readings.
Would you like to test your theory? It's a simple
test. First, be safe and kill the power to your heater.
Install a 100 ohm resistor in series with 1 element
and a new 9 volt battery. Measure the voltage across
the 100 ohm resistor. If the elment is 97 ohms,
you should get about 4.56 volts across the 100 ohm
resistor. If it is 9.7, you should get about 8.2
volts across the 100 ohm resistor.
Finally, re-check the resistance of the element
with your meter.
Ed
>
> Can't be 9.7 ohms as the current would be much too high.
>
>
>
>
> gfretwell@aol.com wrote:
>
>
>
| |
| Don Kelly 2006-06-22, 3:25 am |
| "John C" <johnwcarden@hotmail.com> wrote in message
news:QqydnRJfBPA55wXZnZ2dnUVZ8qKdnZ2d@bt.com...
> resistance is relative.
> You need to include impedance of the total cct in parallel.
>
> p=ir
> 4500=970*I
> I=4500/970
> 4.6amps......
>
> 1/970 etc
>
> total resistance is 323 ohms
>
> more confusion?
-------------------
Since when is p=ir?? Also the apparent resistance of an element is 97 ohms,
not 970.
It appears that the ohmmeter reading is in error but his arithmetic is
correct.
--
Don Kelly dhky@shawcross.ca
remove the X to answer
----------------------------
>
>
> "Palindr?me" <me9@privacy.net> wrote in message
> news:129gt19bm2g4988@corp.supernews.com...
[color=darkred]
>
>
| |
| Don Kelly 2006-06-22, 3:25 am |
| <david.cawkwell@tesco.net> wrote in message
news:1150902744.545353.323530@r2g2000cwb.googlegroups.com...
> No I measured all 3 on the 200ohms scale of the meter.
> All came back with 97 ohms.
>
> I'll go with the idea that the resistance will drop as the elements get
> warm.
>
> Can't be 9.7 ohms as the current would be much too high.
>
----------
An ohmmeter is often unreliable in making measurements on power circuits. At
240V, the effect of some surface film is often negligable (it breaks down
and conducts well ) but at the voltage provided by a meter battery, it is an
important factor adding to the apparent resistance as seen by the ohmmeter.
You may be able to get around this by scratching the terminals of the
element (likely copper) to get bare metal and then using your meter. If the
results still seem off, then measure the current, as indicated below, at
240V.
In addition, the hot resistance will be higher than that of a cold element
so the chances are that the cold element resistance will be less than 38
ohms and rise to 38 ohms after it heats up.
If you do not have an ammeter I suggest that you put a 0.1 ohm, 5 watt ohm
resistor in series with the element and apply 240V to the combination
(measure it), and measure the voltage drop across the resistor. Multiply
this by 10 to get the current.
--
Don Kelly dhky@shawcross.ca
remove the X to answer
----------------------------
>
> gfretwell@aol.com wrote:
>
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